Attempt to insert data into SQLite database

**GarryBrown** · 02-26-2013 08:10 AM

Hello, i am trying to insert data about picture in the Picture table. I do not understand what's going on here, but nothing works. Basically, the database file is created so it's fine. The only issue is inserting new values. I am using method post. Can anyone tell me what am I doing wrong? Thanks in advance.

Creating database file:

Code:

<?

$db=sqlite_open("art.db");



@sqlite_query($db, "DROP TABLE Picture");
@sqlite_query($db, "DROP TABLE Artist");
@sqlite_query($db, "DROP TABLE Sales");
@sqlite_query($db,"CREATE TABLE Picture (ID integer  PRIMARY KEY , name VARCHAR[200], year integer, artistID integer)",$sqliteerror);
@sqlite_query($db,"CREATE TABLE Artist (IDnum integer PRIMARY KEY , aname VARCHAR[10], DOB integer, DOD integer, nationality VARCHAR[200])",$sqliteerror);
@sqlite_query($db,"CREATE TABLE Sales (picID integer PRIMARY KEY , sdate date, price decimal)",$sqliteerror);



sqlite_query($db,"INSERT INTO Picture (ID, name, year, artistID) VALUES ( 1, 'The Hay Wain', 1821, 4 )");
sqlite_query($db,"INSERT INTO Picture (ID, name, year, artistID) VALUES ( 2, 'Salisbury Cathedral from the Meadows', 1831, 4 )");
sqlite_query($db,"INSERT INTO Picture (ID, name, year, artistID) VALUES ( 3, 'A Wheatfield with Cypresses', 1889, 7 )");


sqlite_query($db,"INSERT INTO Artist (IDnum, aname, DOB, DOD, nationality) VALUES ( 4, 'John Constable', 1776, 1837, 'British' )");
sqlite_query($db,"INSERT INTO Artist (IDnum, aname, DOB, DOD, nationality) VALUES ( 7, 'Vincent Van Gogh', 1853, 1890, 'Dutch' )");


sqlite_query($db,"INSERT INTO Sales (picID, sdate, price) VALUES ( 1, 2008-01-01, 23.56 )");
sqlite_query($db,"INSERT INTO Sales (picID, sdate, price) VALUES ( 1, 2008-01-09, 500230.00 )");
sqlite_query($db,"INSERT INTO Sales (picID, sdate, price) VALUES ( 1, 2009-12-15, 87.56 )");




sqlite_close($db);

Html interface for inserting data about picture:

Code:

<html>
<head>
</head>
<body>
<form action="process.php" method="POST">
    <table>
        <tr>
            <td>ID:</td><td><input name="ID"/></td>
        </tr>
        <tr>
            <td>Name:</td><td><input name="Name"/></td>
        </tr>
        <tr>
            <td>Year:</td><td><input name="Year"></textarea></td>
        </tr>
		<tr>
            <td>ArtistID:</td><td><input name="ArtistID"></textarea></td>
        </tr>
        <tr>
            <td></td><td><input type="submit" name="submit" value="Submit"/></td>
        </tr>
    </table>
</form>
</body>
</html>

PHP file inserting the actual data and trying to display it in the table:

Code:

<body>
<?php

//SQLite Database test query

$db=sqlite_open("art.db");

$ID = $_POST["ID"]; 
$name = $_POST["name"]; 
$Year = $_POST["Year"];
$ArtistID = $_POST["ArtistID"];

sqlite_query($db,"INSERT INTO Picture (ID, name, year, artistID) VALUES ( '&ID', '&name', '&Year', '&ArtistID' )");

//now output the data to a simple html table...
echo "<table border=1>";
echo "<tr><td>Id</td><td>Name</td><td>Year</td><td>ArtistID</td></tr>";
while($row=sqlite_fetch_array($result,SQLITE_ASSOC ))
{
	echo "<tr>";
	echo "<td>" . $row['ID'] . "</td>";
	echo "<td>" . $row['name'] . "</td>";
	echo "<td>" . $row['Year'] . "</td>";
        echo "<td>" . $row['ArtistID'] . "</td>";

   echo "</tr>";
}
echo "</table>";


sqlite_close($db);

	
?>

</body>
</html>

**NogDog** · 02-26-2013 08:29 AM

I'd suggest checking the return value from each sqlite_*() call, and if false, output or log debug info, including whatever sqlite_error_string() has to say at that point. Then we can narrow down where the problem(s) is (are).

**GarryBrown** · 02-26-2013 10:56 AM

Thank you for the reply. However, I do not know how to do that cause I am new to all this stuff.

**NogDog** · 02-26-2013 11:02 AM

The quick and dirty way:

PHP Code:


$result = sqlite_query($db,"INSERT INTO Picture (ID, name, year, artistID) VALUES ( '&ID', '&name', '&Year', '&ArtistID' )");
if($result == false) {
    die("Insert failed: ".sqlite_error_string());
}

**OctoberWind** · 02-26-2013 11:13 AM

two things that jumped out at me:

PHP Code:


sqlite_query($db,"INSERT INTO Picture (ID, name, year, artistID) VALUES ( '&ID', '&name', '&Year', '&ArtistID' )");

Should it not be:

PHP Code:


... VALUES ( '$ID', '$name', '$Year', '$ArtistID' )  // dollar, not ampersand

PHP Code:


$name = $_POST["name"]; // based on the form, this should be "Name"

**GarryBrown** · 02-26-2013 04:59 PM

NotDog, thank you very much for your example of error handling. I shall practice this and use for future.
OctoberWind, i've noticed earlier my mistake which was using & instead of $ and corrected it. I haven't noticed, however that in my form it was "Name" with capital N, and this is why the program wasn't working properly.
Now everything works just as it should be !

))))))

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